Lid

What is the weight of the concrete cover (lid) of a round well with a diameter 1.65 m if the thickness of the cover is 9 cm? 1 m³ of concrete weighs 2190 kg.

Final Answer:

m =  421.4 kg

Step-by-step explanation:

$$\begin{gathered} D=1.65\text{m} \\ h=9\text{cm}\to \text{m}=9:100\text{m}=0.09\text{m} \\ \rho =2190{\text{kg/m}}^3 \\ r=D/2=1.65/2=\frac{33}{40}=0.825\text{m} \\ S=\pi \cdot r^2=3.1416\cdot 0.825^2\doteq 2.1382{\text{m}}^2 \\ V=h\cdot S=0.09\cdot 2.1382\doteq 0.1924{\text{m}}^3 \\ m=\rho \cdot V=2190\cdot 0.1924=421.4\text{kg} \end{gathered}$$

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